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Prove that if $h$ is infinitely differentiable in a neighborhood of $0$, then the kth derivative evaluated at 0 is


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What have you tried? Some people consider it rule when they see something like "Prove this". – glebovg Nov 23 '12 at 21:58
I really want to see the answer to this. – 000 Nov 23 '12 at 22:01
up vote 1 down vote accepted

Define $$\Delta_{\delta}^{(n)}(h(x)) = \sum_{k=0}^{n}(-1)^{(n-k)} \dbinom{n}k h(x+k\delta)$$ We will then prove that $$\dfrac{d^n h(x)}{dx^n} = \lim_{\delta \to 0}\dfrac{\Delta_{\delta}^{(n)}(h(x))}{\delta^n}$$ If we prove this, then your result follows immediately by plugging in $x=0$. The proof not surprisingly goes by induction on $n$.

Clearly, for $n=1$ it is true since $$\Delta_{\delta}^{(1)}(h(x)) = \sum_{k=0}^{1}(-1)^{(1-k)} \dbinom{1}k h(x+k\delta) = -h(x) + h(x+\delta)$$ Hence, $$\lim_{\delta \to 0}\dfrac{\Delta_{\delta}^{(1)}(h(x))}{\delta} = \lim_{\delta \to 0} \dfrac{h(x+ \delta) - h(x)}{\delta} = \dfrac{dh(x)}{dx}$$ Assume it is true for $n = m$. We need to prove it for $n=m+1$. Note that $$\Delta_{\delta}^{(m+1)}(h(x)) = \sum_{k=0}^{m+1}(-1)^{(m+1-k)} \dbinom{m+1}k h(x+k\delta)$$ Consider $\Delta_{\delta}^{(m)}(h(x+\delta)) - \Delta_{\delta}^{(m)}(h(x))$. \begin{align} \Delta_{\delta}^{(m)}(h(x+\delta)) - \Delta_{\delta}^{(m)}(h(x))& = \sum_{k=0}^{m}(-1)^{(m-k)} \dbinom{m}k \left(h(x+\delta+k\delta) - h(x+k\delta) \right)\\ & = (-1)^{m+1} h(x) + h(x+(m+1) \delta) + g_m(x,\delta) \end{align} where \begin{align} g_m(x,\delta) & = \sum_{k=0}^{m-1} (-1)^{(m-k)} \dbinom{m}k h(x+\delta+k\delta) - \sum_{k=1}^{m} (-1)^{(m-k)} \dbinom{m}k h(x+k\delta)\\ & = \sum_{k=1}^{m} (-1)^{(m-k+1)} \dbinom{m}{k-1} h(x+k\delta) - \sum_{k=1}^{m} (-1)^{(m-k)} \dbinom{m}k h(x+k\delta)\\ & = \sum_{k=1}^{m} (-1)^{(m-k+1)} \dbinom{m+1}{k} h(x+k\delta) \left(\because \dbinom{m}{k} + \dbinom{m}{k-1} = \dbinom{m+1}{k} \right) \end{align} Hence, we have that \begin{align} \Delta_{\delta}^{(m)}(h(x+\delta)) - \Delta_{\delta}^{(m)}(h(x))& = \sum_{k=0}^{m}(-1)^{(m-k)} \dbinom{m}k \left(h(x+\delta+k\delta) - h(x+k\delta) \right)\\ & = (-1)^{m+1} h(x) + h(x+(m+1) \delta)\\ & + \sum_{k=1}^{m} (-1)^{(m-k+1)} \dbinom{m+1}{k} h(x+k\delta)\\ & = \sum_{k=0}^{m+1} (-1)^{(m-k+1)} \dbinom{m+1}{k} h(x+k\delta)\\ & = \Delta_{\delta}^{(m+1)}(h(x)) \end{align} Hence, now we have that $$\dfrac{d^{(m+1)}h}{dx^{(m+1)}} = \lim_{\delta \to 0}\dfrac{\dfrac{d^{(m+1)}h(x+\delta)}{dx^{(m)}} - \dfrac{d^{(m)}h(x)}{dx^{(m)}}}{\delta} = \lim_{\delta \to 0}\dfrac{\lim_{\delta_1 \to 0}\dfrac{\Delta_{\delta_1}^{(m)}(h(x+\delta))}{\delta_1^m} - \lim_{\delta_2 \to 0}\dfrac{\Delta_{\delta_2}^{(m)}(h(x))}{\delta_2^m}}{\delta}$$ Since all limits behave nicely, we can afford to choose $\delta_1 = \delta_2 = \delta$, to give $$\dfrac{d^{(m+1)}h}{dx^{(m+1)}} = \lim_{\delta \to 0}\dfrac{\Delta_{\delta}^{(m)}(h(x+\delta)) - \Delta_{\delta}^{(m)}(h(x))}{\delta^{m+1}} = \lim_{\delta \to 0}\dfrac{\Delta_{\delta}^{(m+1)}(h(x))}{\delta^{m+1}}$$ which proves our claim that $$\dfrac{d^n h(x)}{dx^n} = \lim_{\delta \to 0}\dfrac{\Delta_{\delta}^{(n)}(h(x))}{\delta^n}$$ where $$\Delta_{\delta}^{(n)}(h(x)) = \sum_{k=0}^{n}(-1)^{(n-k)} \dbinom{n}k h(x+k\delta)$$

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Taylor's formula allows us to write $$h(jt)=\sum_{l=0}^n\frac{(jt)^l}{l!}h^{(l)}(0)+t^n\varepsilon_j(t),$$ where $\lim_{t\to 0}\varepsilon_j(t)=0$. This gives, defining $\varepsilon(t):=\sum_{j=0}^n\varepsilon_j(t)$, \begin{align} \sum_{j=0}^n\binom nj(-1)^{n-j}h(jt)&=\sum_{j,l=0}^n\binom nj\frac{(-1)^{n-j}(jt)^l}{l!}h^{(l)}(0)+t^n\varepsilon(t)\\ &=\sum_{l=0}^n\frac{t^l}{l!}\sum_{j=0}^n\binom nj j^l(-1)^{n-j}+t^n\varepsilon(t). \end{align} Let $c_l:=\sum_{j=0}^n\binom nj j^l(-1)^{n-j}$ for $0\leqslant l\leqslant n$ and $p(s)=\sum_{j=0}^n\binom nje^{sj}(-1)^{n-j}=(e^s-1)^n$. Then $c_l=p^{(l)}(0)$. If we take the derivative less than $n-1$ times evaluated at $0$, we will get $0$.

To conclude, we need to show that $c_n=n!$. A proof can be found here.

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